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Math Help - In need of direction

  1. #1
    Member M670's Avatar
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    In need of direction

    Suppose
    and is negative. Here are some small variations on the previous problems:






    Can someone please point me in a direction towards solving this problem, I am not sure what they are asking?
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  2. #2
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    Re: In need of direction

    \cos{u} > 0 and \sin{u} < 0 indicates u is a quad IV angle.

    \cos{u} = \frac{5}{13} \, , \, \sin{u} = -\frac{12}{13}


    to calculate the rest, use the difference identity for sine and cosine ...

    \sin(a - b) = \sin{a}\cos{b} - \cos{a}\sin{b}

    \cos(a-b) = \cos{a}\cos{b} + \sin{a}\sin{b}
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  3. #3
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    Re: In need of direction

    You can also note that a right triangle with hypotenuse of length 13 and one leg of length 5 has cosine of the angle between those two side 5/13. What is the length of the other leg? And from that, what is the sine of the angle? This is treating the angle as between 0 and \pi/2. Use the quadrant information to find the sign.
    Last edited by HallsofIvy; November 12th 2012 at 04:56 PM.
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  4. #4
    Member M670's Avatar
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    Re: In need of direction

    Quote Originally Posted by HallsofIvy View Post
    You can also note that a right triangle with hypotenuse of length 13 and one leg of length 5. What is the length of the other leg?
    12 ahh I see how you are looking at it...
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  5. #5
    Member M670's Avatar
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    Re: In need of direction

    so then for sin(u-pi) it should look like this =sin(u)cos(pi)-cos(u)sin(pi)
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  6. #6
    Member M670's Avatar
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    Re: In need of direction

    or would it be sin(-(12/13)-pi)= sin-(12/13)cos(pi)-cos-(12/13)sin(pi)?
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